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Math Help - Special quadrangle with 135 and 90

  1. #1
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    Special quadrangle with 135 and 90

    Hi,
    I hope you guys can help me with the following problem:
    I have a quadrangle ABCD. <ADC = 135, <ABC = 90, [AB] = [BC]
    That's all I know, but now I have to prove that [BD] = [AB] = [BC]
    I tried some things with equations but it doesn't really work.

    Thanks for your help!
    Btw: I'm from Austria and I'm not a native speaker, so please feel free to correct my linguistic mistakes.
    Last edited by jalt; February 13th 2013 at 11:59 AM.
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  2. #2
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    Re: Special quadrangle with 135 and 90

    At first I didn't even believe it, but dynamic geometry convinced me otherwise. Here's a solution:

    Special quadrangle with 135° and 90°-mhfgeometry7.png
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  3. #3
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    Re: Special quadrangle with 135 and 90

    Hi,
    thanks a lot for your reply and your solution. I know this might sound strange but my teacher told me that there was a solution without any trigonometry.
    Unfortunately he isn't gonna tell me until I find it out.
    I think I tried everything but it seems to be impossible to me. Do you have another idea or can you at least tell me whether it's possible or not?
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  4. #4
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    Re: Special quadrangle with 135 and 90

    Jalt

    your teacher is right
    you do not need all the complicated answers..after all the beuty of Mathematics lies on one word SIMPLICITY.
    Anyway the quadrelateral you mention is part of a regular octagone with B as center of the circle... check it.....as simple as such

    MINOAS
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  5. #5
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    Re: Special quadrangle with 135 and 90

    I'm afraid I disagree with the previous answer. A regular octagon has all of its sides equal, but there's no restriction on your quadrilateral that AD = DC.

    Back to the problem. Since the result is true, the circumcenter of triangle ACD mus be B; i.e. the intersection of the perpendicular bisectors of AD and DC must be B. I can see no reason why this is true. There's probably a simple geometric argument, but I can't see it.
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  6. #6
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    Re: Special quadrangle with 135 and 90

    John

    construct a regular octagone with B as the center of circle.and BA=BD=BC as the radii of the circle .The sides AD=DC .The angle ABC=90 degrees and angle abd=135 degrees.
    it is very simple try it.
    Minoas
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  7. #7
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    Re: Special quadrangle with 135 and 90

    Minoanman,

    What you say about constructing a regular octagon is certainly true, but it does not answer the original question. The sides AD and DC are not necessarily equal. See the figure in my initial response.
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  8. #8
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    Re: Special quadrangle with 135 and 90

    construct a right angle at B with equal legs BA and BC.Swing an arc from C to A.Mark a point D anywhere on the arc. Prove that the angle ADC is always 135 deg It is simple
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  9. #9
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    Re: Special quadrangle with 135 and 90

    JOHNG
    Construct a regular octagone with B as the center of circle and BA,BC BD RADII . THE QUADRILATERAL ABCD IS ONE OF THE TRILLIONS YOU CAN GET WITH THE SAME PROPERTY ONCE YOU SLIDE THE VERTEX D OVER THE ARC AD. TRY IT IT IS SIMPLE. I AGREE WITH BJHOPPER IT IS SIMPLE.

    MINOAS
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  10. #10
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    Re: Special quadrangle with 135 and 90

    I agree with bjhopper that his construction always produces a quadrilateral as specified originally -- it's nothing more than an application of the inscribed angle theorem. I'm not being obtuse on purpose, but why is every quadrilateral, as originally described, an instance of this construction? It seems to me that you need to know that length BD is the same as BA and BC (of course this is true, but this was the original question).

    minoanman: I think my source of confusion with you is that you keep saying regular octagon. As defined everywhere, a regular polygon must have sides of equal length. In particular, a regular octagon has equal sides.
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  11. #11
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    Re: Special quadrangle with 135 and 90

    Geometry fascinates me, but I've always known my skills at geometry are poor. The missing item in my information bank was:
    Given an obtuse triangle ADC with obtuse angle \theta at D, the circumcenter of ADC is the unique point P on the perpendicular bisector of AC which satisfies:
    1. P is in that half plane determined by line AC which does not contain D.
    2. The angle APC is 2\pi-2\theta.

    So the proof as suggested by bjhopper is absolutely correct. There's nothing magic about 135 degrees, it can be replaced by any \theta and the angle ABC being 2\pi-2\theta.

    MY ignorance caused all the hubbub. Sorry.
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  12. #12
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    Re: Special quadrangle with 135 and 90

    Hi,
    I am sorry, I was on vacation the last week.
    Unfortunately I still haven't understood your solution.

    I have an arc with the center B and the radius AB and now I claim that <ADC is always 135 if D is on the arc.
    Well that's true but how can I prove it?

    @johng What angle is pi?
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  13. #13
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    Re: Special quadrangle with 135 and 90

    Hi jalt,
    See post 8.
    Triangle ABD and DBC are isosceles.Sides are radii
    Angle BAD=BDA =a Angle BDC =BCD =b Property of Isosceles triangles
    360 =90 +135 +2a +2b Property of quads
    270 = 2a+2b
    a+b = 135 angle D.D can be anywhere on the arc CA
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  14. #14
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    Re: Special quadrangle with 135 and 90

    Thanks a lot!
    I am still just a little confused, because you proved that D can be anywhere on the arc CA, but you didn't prove that D has to be on the arc.
    Why isn't it possible that there is a point outside of the arc that makes <ADC = 135?
    There's probably a very easy solution but I don't see it.
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  15. #15
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    Re: Special quadrangle with 135 and 90

    Jalt,

    I hope the attached diagram and explanation helps:

    Special quadrangle with 135° and 90°-mhfgeometry13.png
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